Chapter 2: Problem 6
Suppose that \((b-a)^{2}-4 a c>0 .\) a) Find a formula for the general solution of \(a x^{2} y^{\prime \prime}+b x y^{\prime}+c y=0 .\) Hint: Try \(y=x^{r}\) and find a formula for \(r\). b) What happens when \((b-a)^{2}-4 a c=0\) or \((b-a)^{2}-4 a c<0 ?\) We will revisit the case when \((b-a)^{2}-4 a c<0\) later.
Short Answer
Step by step solution
Substitute Trial Solution
Compute Derivatives
Substitute Derivatives into DE
Simplify Equation
Solve for r
Write General Solution
Analyze the Discriminant
Special Cases - Zero Discriminant
Special Cases - Negative Discriminant
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
General Solution
- \( y = x^r \)
- \( y' = r x^{r-1} \)
- \( y'' = r(r-1) x^{r-2} \)
Quadratic Equation
Discriminant Analysis
- If \( (b-a)^2 - 4ac > 0 \), the roots \( r_1 \) and \( r_2 \) are real and distinct.
- If \( (b-a)^2 - 4ac = 0 \), the equation has a double root, leading to a solution form of \( y(x) = C_1 x^{r} + C_2 x^{r} \ln{x} \).
- If \( (b-a)^2 - 4ac < 0 \), the roots are complex and thus conjugates, necessitating further exploration into complex solutions.
Complex Conjugates
- A complex solution might be presented as \( y(x) = x^\alpha (C_1 \cos(\beta \ln{x}) + C_2 \sin(\beta \ln{x})) \).